Answer:
20,861.65 mi/h²
Explanation:
We convert 4,470,000 pound-mass [lbm], to kg. Since 2.205 lbm = 1 kg, then 4,470,000 lbm = 4,470,000 lbm × 1 kg/2.205 = 2,027,210.88 kg
Since Force , F = ma where m = mass and a = acceleration, and our force of thrust , F = 5.25 MN = 5,250,000 N and or mass = mass of spacecraft = 2,027,210.88 kg, we then find the acceleration, a.
a = F/m = 5,250,000 N/2,027,210.88 kg = 2.59 m/s².
We now convert this acceleration into miles per hour. Since 1 mile = 1609 meters and 60 × 60 s = 1 hour ⇒ 3600 s = 1 hour, Our conversion factor for meter to mile is 1 mile/1609 m and that for second to hour is 3600 s/1 hour. We square the conversion factor for the time so we have (3600 s/1 hour)².
Multiplying both conversion factors with our acceleration, we have
a = 2.59 m/s²
= 2.59 m/s² × 1 mile/1609 m × (3600 s/1 hour)²
= 33566440/1609 miles/hour²
= 20,861.65 mi/h²
= 20,861.65 miles per hour squared
A particle with charge q = +5.00 C initially moves at v = (1.00 î + 7.00 ĵ ) m/s. If it encounters a magnetic field B = 10 k, find the magnetic force vector on the particle.
Given :
Force on object moving with constant velocity due to magnetic field is given by :
[tex]\vec{F}=q\vec{v}\times \vec{B}\\\\\vec{F}=-5[ ( \hat{i} + 7\hat{j}) \times ( 10\hat{k})]\\\\\vec{F}=-5[ 10(-\hat{j})+70(\hat{i})]\\\\\vec{F}=50\hat{j}-350\hat{i}\ N[/tex]
Now,
[tex]F = \sqrt{50^2+350^2}\\\\F=353.55\ N[/tex]
Hence, this is the required solution.
A particle with charge q = +5.00 C that is moving with some velocity, then the magnetic force vector on the particle will be 353.55 N.
What is a vector?The phrase "vector" in physics and mathematics is used popularly to refer to some values that cannot be stated by a simple integer (a scalar) or to certain elements of high - dimensional space.
For things like displacements, forces, and velocity can have both a magnitude and direction, vectors were initially introduced in geometry and physics. Similar to how lengths, masses, and time are represented by real numbers, these quantities were represented by geometric vectors.
In some contexts, tuples—finite sequences of numbers with a definite length—are sometimes referred to as vectors.
From the data given in the question,
F = qv × B
F = -5 [(î +7 ĵ) ×(10 k)]
F = 50 ĵ - 350 î N
Now, calculate the force,
F = √50² + 350²
F = 353.55 N.
To know more about Vector:
https://brainly.com/question/13322477
#SPJ5
A 80-kg person stands at rest on a scale while pulling vertically downwards on a rope that is above them. Use g = 9.80 m/s2.
With what magnitude force must the tension in the rope be pulling on the person so that the scale reads 25% of the persons weight?
Answer:
The tension in the rope is 588 N
Explanation:
I have attached a free body diagram of the problem.
First we calculate the force applied downwards by the person's weight:
[tex]F_{weight} =m*a = 80 kg * 9.8 \frac{m}{s^{2}} = 784 N[/tex]
If the scale reads a 25% of the person' weight:
[tex]F_{Scale} = F_{weight}*0.25=784N*0.25=196N[/tex]
which is the same value (but opposite) to the effective weight of the persons ([tex]F_{Person}=-F_{Scale}[/tex])
The other 75% of the total weight is the force of the rope pulling the person upwards.
[tex]F_{Pull}= F_{weight}*0.75=784N*0.75=588N[/tex]
To verify, we can use 1st Newton's law
∑F = 0
[tex]F_{Rope}-F_{Pull}+F_{Person}-F_{Scale}=0 \\588N-588N+196N-196N=0[/tex]
4. A 5.0 kg block moving to the right at 12.0 m/sec collides with a 4.0 kg block moving to the right at 2 m/sec. If the 4.0 kg moves to the right at 10 m/sec after the collision, what will be the speed of the 5.0 kg block? How much energy was lost during the collision?
Answer:
Energy lost = 55.28Joules
Explanation:
Using the law of momentum:
m1u1+m2u2 = m1v1+m2v2
m1 and m2 are the masses of the objects
u1 and u2 are the initial velocities.
v1 and v2 are final velocities
Given
m1 = 5kg
m2 = 4kg
u1 = 12m/s
u2 = 2m/s
v1 = 10m/s
v2 = ?
Substitute
5(12)+4(2)=4(10)+5v2
60+8 = 40+5v2
68-40 = 5v2
28 = 5v2
v2 = 28/5
v2 = 5.6m/sec
Hence the speed of the 5kg block is 5.6m/s
Energy lost = Kinetic energy after collision - kinetic energy before collision
KE after collision = 1/2m1v1²+1/2m2v2²
KE after collision = 1/2(5)10²+1/2(4)(5.6)²
= 250+62.72
= 312.72Joules
KE before collision = 1/2m1u1²+1/2m2u2²
KE before collision = 1/2(5)12²+1/2(4)(2)²
= 360+8
= 368Joules
Energy lost = 368-312.72
Energy lost = 55.28Joules
A Ferris wheel rotates at an angular velocity of 0.25 rad/s. Starting from rest, it reaches its operating speed with an average angular acceleration of 0.027 rad/s2. How long does it take the wheel to come up to operating speed?
Answer:
t = 9.25 s
Explanation:
Given that,
Initial angular velocity, [tex]\omega_o=0[/tex] (at rest)
Final angular velocity, [tex]\omega_o=0.25\ rad/s[/tex]
Angular acceleration, [tex]\alpha =0.027\ rad/s^2[/tex]
We need to find the time it take the wheel to come up to operating speed. We know that the angular acceleration in terms of angular speed is given by :
[tex]\alpha =\dfrac{\omega_f-\omega_o}{t}\\\\t=\dfrac{\omega_f-\omega_o}{\alpha }\\\\t=\dfrac{0.25-0}{0.027}\\\\t=9.25\ s[/tex]
So, it will reach up to the operating speed in 9.25 s.
will give brainliest pls help
Answer:
a and c
Explanation:
A race car accelerates from rest with a displacement of 25.0 m. If the
acceleration is 1.2 m/s2, what is the car's final velocity?
Answer:
Vf = Vi + at
t = Vf - Vi = 0.0m/s - 1.75 m/s
--------- = ------------------------- = 8.8s
a -0.20 m/s(squared)
Explanation:
This is what I got I am pretty sure it's correct but correct me if I'm wrong : )
A 3.00-kg block starts from rest at the top of a 35.5° incline and slides 2.00 m down the incline in 1.45 s. (a) Find the acceleration of the block.
Answer:
0.344 m/s^2
Explanation:
The first step is to calculate V
V= distance /time
= 2/1.45
= 1.379 m/s
Therefore the acceleration of the block can be calculated as follows
a= V - U/2d
= 1.379 - 0/2×2
= 1.379/4
= 0.344 m/s^2
The microwave background radiation is observed at a wavelength of 1.9 millimeters. The red shift of these photons is observed to be raoughly 1100. What would be the emitted wavelength?
Answer:
1730nm
Explanation:
Observed wavelength λ= 1.9 millimeters
Red shift of photons = 1100
Emitted wavelength = λo
Red shift = λ - λo/ λo = 1100
= [(1.9x10^-3 - λo)/ λo] = 1100
= 1.9x10^-3 - λo = 1100 λo
= 1.9x10^-3 = 1100 λo + λo
= 1.9x10^-3 = 1101 λo
We divide through by 1101 to get the value of lambda
1.9x10^-3/1101 = λo
λo = 0.000001726
λo = 1726nm
This is approximately
λo = 1730nm
Thank you!
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 4.40 rev/s; 60.0 revolutions later, its angular speed is 15.0 rev/s. Calculate (a) the angular acceleration (rev/s2), (b) the time required to complete the 60.0 revolutions, (c) the time required to reach the 4.40 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 4.40 rev/s angular speed.
Answer:
The answer is below
Explanation:
a) Using the formula:
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=15\ rev/s,w_o=initial\ velocity=4.4\ rev/s, \\\theta=distance=60\ rev\\\\Substituting:\\\\15^2=4.4^2+2(60)\alpha\\\\2(60)\alpha=15^2-4.4^2\\\\2(60)\alpha=205.64\\\\\alpha=1.71\ rev/s^2[/tex]
b) The disk is initially at rest. Using the formula:
[tex]\theta=\omega_ot+\frac{1}{2}\alpha t^2 \\\\but\ \omega_o=0(rest), \theta=60\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\60=0+\frac{1}{2}(1.71) t^2\\\\t=8.4\ s[/tex]
c)
[tex]\omega=\omega_o+\alpha t \\\\but\ \omega_o=0(rest), \omega=4.4 \ rev/s\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\4.4=1.71t\\\\t=2.6\ s[/tex]
d)
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=4.4\ rev/s,w_o=initial\ velocity=0\ rev/s, \\\alpha=1.71\ rev/s^2\\\\Substituting:\\\\4.4^2=0+2(1.71)\theta\\\\19.36=3.42\theta\\\\\theta=5.66\ rev[/tex]
ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?
Answer:
A)90°
Explanation:
ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?
(a) What is the angle between and the coil's magnetic dipole moment.
The angle between and the coil's magnetic dipole moment is 90°, because it is perpendicular to the field since there is maximum torgue.
(b) What is the number of turns in the coil?
Explain the differences between getting burned by fire and getting burned by dry ice?
Need it ASAP
Question 2: If a runner travels 50 m in 5 s, his acceleration is
1 point
250 m/s/s
50 m/s/s
10 m/s/s
2 m/s/s
A roller coaster stops at the top of a hill. What force brings the roller coaster back down to the ground?
Answer:
Simple enough that it seems like a trick question: gravity.
State two function of the mass M placed on the metre rule
correct answer is mass media is correct answer OK
A dentist’s drill starts from rest. After 7.28 s of constant angular acceleration it turns at a rate of 16740 rev/min. Find the drill’s angular acceleration. Answer in units of rad/s 2 .
Answer:
[tex]\alpha =240.79\ rad/s^2[/tex]
Explanation:
Given that,
Initial angular velocity, [tex]\omega_i=0[/tex]
Final angular velocity, [tex]\omega_f=16740\ rev/min = 1753\ rad/s[/tex]
We need to find the angular acceleration of the drill. It is given by the formula as follows :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{1753-0}{7.28}\\\\\alpha =240.79\ rad/s^2[/tex]
So, the angular acceleration of the drill is [tex]240.79\ rad/s^2[/tex].
A 1500 kg car hits a haystack at 8m/s and comes to a stop after 0.6 seconds
A)what is the change in momentum of the car ?
B) what is the impulse that acts on the car ?
C) what is the average force that acts on the car during the collision?
D) what is the work done by the haystack in stopping the car ?
Answer:
Kindly check explanation
Explanation:
Change in momentum:
m(v - u)
M = mass = 1500kg
v = final velocity = 0 m/s
u = initial velocity = 8m/s
Momentum = 1500(8 - 0)
Momentum = 1500 * 8
Momentum = 12000 kgm/s
B.) impulse that acts on the car
Impulse = Force * time
Impulse = change in momentum with time
Impulse = m(v - u)
Hence, impulse = 12000Ns
C.) Average force action on the car during collision
Impulse = Force * time
12000 = force * 0.6
12000/ 0.6 = force
Force = 20000 N
D.) Workdone by haystack in stopping the car :
Workdone = Force * Distance
Distance = speed * time
Distance = 8 * 0.6
Distance = 4.8m
Hence,
Workdone = 20,000 * 4.8 = 96000 J
1. a. Describe the type of energy that is observed when a musician's finger plucks a guitar string. does it give your ear kenetic or potential energy .
Answer:
kinetic
Explanation:
Answer:
kinetic energy.
Explanation:
A pair of equal-length vectors at right angles to each other have a resultant. If the angle between the vectors is less than 90°, their resultant is
Answer:
Larger in magnitude
Explanation:
As the vectors get closer reducing the 90 degree angle, based on the "parallelogram rule" for vector addition, the magnitude of the resultant vector gets larger than when they are forming a 90 degree angle.
A tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet. The coefficients of static and kinetic friction are 0.6 and 0.4, respectively. The worker pushes with a horizontal force of 490 N. The frictional force exerted by the surface is
Answer:
The friction force exerted by the surface is 490 newtons.
Explanation:
From Physics, we remember that static friction force ([tex]f_{s}[/tex]), measured in newtons, for a particle on a horizontal surface is represented by the following inequation:
[tex]f_{s} \leq \mu_{s}\cdot m \cdot g[/tex] (Eq. 1)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If [tex]f_{s} = \mu_{s}\cdot m \cdot g[/tex], then crate will experiment an imminent motion. The maximum static friction force is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 100\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f_{s} = (0.6)\cdot (100\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f_{s} = 588.42\,N[/tex]
From Newton's Laws we get that current force of friction as reaction to the pulling force done by the worker on the crate is:
[tex]\Sigma F = F-f = 0[/tex]
[tex]f = F[/tex] (Eq. 2)
Where:
[tex]F[/tex] - Horizontal force done by the worker, measured in newtons.
[tex]f[/tex] - Static friction force, measured in newtons.
If [tex]F = 490\,N[/tex], then the static friction force exerted by the surface is:
[tex]f = 490\,N[/tex]
Given that [tex]f < f_{s}[/tex], the crate does not change its state of motion. The friction force exerted by the surface is 490 newtons.
How much is the spring stretched, in meters, by an object with a mass of 0.49 kg that is hanging from the spring at rest
Answer:
1.6m
Explanation:
Using the spring equation, which is as follows:
F = Kx
Where; F = force applied on the spring (N)
K = spring constant (3kg/s²)
x = extension of spring (m)
However, Force applied by object = mass × acceleration (9.8m/s²)
F = 0.49kg × 9.8m/s²
F = 4.802N
Since, F = 4.802N
F = Kx
4.802 = 3 × x
4.802 = 3x
x = 4.802/3
x = 1.6006
x = 1.6m
The spring is stretched i.e. the extension of the spring, by 1.6m
A student throws a water bottle upward into the air. It takes 1.2 seconds for the bottle to reach it's
maximum height. How much time will it take for the water bottle to return to the student's hand
(from its maximum height)?
Answer:
1.2 seconds
Explanation:
Formula for total time of flight in projectile is; T = 2u/g
Now, for the time to reach maximum height, it is gotten from Newton's first equation of motion;
v = u + gt
t is time taken to reach maximum height.
But in this case, g will be negative since motion is against gravity. Final velocity will be 0 m/s at max height.
Thus;
0 = u - gt
u = gt
t = u/g
Putting t for u/g in the equation of total time of flight, we have;
T = 2t
We are given t = 1.2
Thus, T = 2 × 1.2 = 2.4
Since total time of flight is 2.4 s and time to get to maximum height is 1.2 s, then it means time to fall from maximum height back to the students hand is; 2.4 - 1.2 = 1.2 s
If the ball is released from rest at a height of 0.63 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track
Answer:
When the ball is on the frictionless side of the track , the angular speed is 89.7 rad/s.
Explanation:
Consider the ball is a solid sphere of radius 3.8 cm and mass 0.14 kg .
Given , mass, m=0.14 kg
Ball is released from rest at a height of, h= 0.83 m
Solid sphere of radius, R = 3.8 cm
=0.038 m
From the conservation of energy
ΔK = ΔU
[tex]\frac{1}{2}mv^2 +\frac{1}{2} I\omega^2=mgh[/tex]
Here , [tex]I=\frac{2}{5} MR^2 , v= R \omega[/tex]
[tex]\frac{1}{2}mv^2\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R^2} )=mgh[/tex]
[tex]\frac{1}{2} [v^2+\frac{2}{5}v^2]= gh[/tex]
[tex]\frac{7}{10} v^2=gh[/tex]
[tex]0.7v^2=gh[/tex]
v=[tex]\sqrt{[gh/(0.7)][/tex]
=[tex]\sqrt{ [(9.8 m/s^2)(0.83 m) / (0.7) ][/tex]
= 3.408 m/s
Hence, angular speed when it is on the frictionless side of the track,
[tex]\omega=\frac{v}{R}[/tex]
= (3.408 m/s)/(0.038 m)
[tex]\omega[/tex] = 89.7 rad/s
Hence , the angular speed is 89.7 rad/s
1.A large beach ball weighs 4.0 N. One person pushes it with a force of 7.0 N due South while another person pushes it 5.0 N due East. Find the acceleration on the beach ball.
2. What is the weight of a 70 kg astronaut on the earth, on the moon, (g=1.6 m/s2), on Venus (g = 18.7 m/s2) and in outer space traveling at a constant velocity
Answer:
1. force applied southward = -4 j
force applied eastward = 5 i
total force applied = 5i - 4j
magnitude of total force applied = √(5)²+(-4)²
magnitude of total force applied = √25 + 16 = √41
magnitude of total force applied = 6.4N
But the beach ball also weighs 4 N,
which means that a force of 4N is required to overcome the inertia of the ball
Net force applied on the ball = total force applied - force applied by inertia
Net force applied on the ball = 6.4 - 4
Net force applied on the ball = 2.4 N
Mass of the ball:
Mass of the ball = weight of the ball / gravitational constant
Mass of the ball = 4 / 9.8 = 0.4 kg
Acceleration of Ball:
from newton's second law of motion:
F = ma
replacing the variables
2.4 = 0.4 * a (where a is the acceleration of the ball)
a = 2.4/0.4
a = 6 m/s²
2. Mass of astronaut = 70 kg
Weight of Earth:
Weight = Mass * acceleration due to gravity
Weight = 70 * 9.8
Weight = 686 N
Weight on Moon:
Weight = Mass * acceleration due to gravity
Weight = 70 * 1.6 (we are given that g = 1.6 on moon)
Weight = 112 N
Weight on Venus:
Weight = Mass * acceleration due to gravity
Weight = 70 * 18.7 (we are given that g = 18.7 on Venus)
Weight = 1309 N
The speed of a 4.0-N hockey puck, sliding across a level ice surface, decreases at the rate of 2 m/s 2. The coefficient of kinetic friction between the puck and ice is:
Answer:
μ = 0.2041
Explanation:
Given
[tex]a = -2m/s^2[/tex]
Required
Determine the coefficient of friction (μ)
The acceleration is negative because it is decreasing.
Solving further, we have that:
[tex]F_f =[/tex] μmg
Where
μ = Coefficient of kinetic friction
[tex]F_f = ma[/tex]
So, we have:
ma = μmg
Divide both sides by m
a = μg
Substitute values for a and g
2 = μ * 9.8
Solve for μ
μ = 2/9.8
μ = 0.2041
How much force (in N) is exerted on one side of an 35.3 cm by 50.0 cm sheet of paper by the atmosphere
Answer:
A force of 1.788 newtons is exerted on one side of the sheet of paper by the atmosphere.
Explanation:
From definition of pressure, we get that force exerted by the atmosphere ([tex]F[/tex]), measured in newtons, on one side of the sheet of paper is given by the expression:
[tex]F = P_{atm}\cdot w\cdot l[/tex] (Eq. 1)
Where:
[tex]P_{atm}[/tex] - Atmospheric pressure, measured in pascals.
[tex]w[/tex] - Width of the sheet of paper, measured in meters.
[tex]l[/tex] - Length of the sheet of paper, measured in meters.
If we know that [tex]P_{atm} = 101325\,Pa[/tex], [tex]w = 0.353\,m[/tex] and [tex]l = 0.05\,m[/tex], then the force exerted on one side of the sheet of paper is:
[tex]F = (101325\,Pa)\cdot (0.353\,m)\cdot (0.05\,m)[/tex]
[tex]F = 1.788\,N[/tex]
A force of 1.788 newtons is exerted on one side of the sheet of paper by the atmosphere.
9. A 50 kg halfback is in the process of making a turn on a football field.
The halfback makes 1/4 of a turn with a radius of 15 meters in 2.1 seconds
before being tackled. What is the net force acting on the halfback before
he is tackled?
Answer:
Net force = 419.5N
Explanation:
Given the following data;
Mass = 50kg
Radius = 15m
Time = 2.1 secs
Turns = 1/4 = 0.25
In order to find the net force, we would first of all solve for the speed and acceleration of the halfback.
To find speed;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the equation;
[tex]Speed = \frac{distance}{time}[/tex]
But the distance traveled is given by the circumference of a circle = [tex] 2\pi r[/tex]
Since he covered 1/4th of a turn;
[tex] Distance = 0.25 * 2 \pi *r[/tex]
Substituting into the equation;
[tex] Speed, v = \frac {(0.25*2*3.142 * 15)}{2.1}[/tex]
[tex] Speed, v = \frac {23.565}{2.1}[/tex]
Speed, v = 11.22m/s
To find acceleration;
[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]
Where, v = 11.22m/s and r = 15m
Substituting into the equation, we have;
[tex] Acceleration, a = \frac {11.22^{2}}{15}[/tex]
[tex] Acceleration, a = \frac {125.8884}{15}[/tex]
Acceleration, a = 8.39m/s²
To find the net force;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.m represents the mass of an object.a represents acceleration.Substituting into the equation, we have;
[tex] F = 50 * 8.39[/tex]
F = 419.5N
Therefore, the net force acting upon the halfback is 419.5 Newton.
differentiate between speed and velocity
Explanation:
Speed - The rate at which something moves
Velocity - The speed of something in a specific direction
Velocity is kind of a specific type of speed.
When an unknown resistance RxRx is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting R3R3 to be 2500ΩΩ. What is RxRx if R2R1=0.625R2R1=0.625?
Answer:
Rₓ = 1562.5 Ω
Explanation:
The formula for the wheat stone bridge in balanced condition is given as follows:
R₁/R₂ = R₃/Rₓ
where,
Rₓ = Unknown Resistance = ?
R₃ = 2500 Ω
R₂/R₁ = 0.625
R₁/R₂ = 1/0.625 = 1.6
Therefore,
1.6 = 2500 Ω/Rₓ
Rₓ = 2500 Ω/1.6
Rₓ = 1562.5 Ω
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.30 mm high. Assume it starts from rest and rolls without slipping.
Answer:
9.77 m/s
Explanation:
Since the cylinder is rolling, it will have both translational and rotational kinetic energy.
From conservation of energy, its initial mechanical energy equals its final mechanical energy.
So, K' + U' = K + U where K and U are the initial kinetic and potential energies respectively and K' and U' are the final kinetic and potential energies respectively.
So, Since the cylinder starts from rest, its initial kinetic energy K = 0, its initial potential energy = mgh where m = mass of cylinder, g = acceleration due to gravity = 9.8 m/s² and h = height of incline = 7.30 mm = 7.3 × 10⁻³ m . Its final kinetic energy K' = 1/2Iω² + 1/2mv² where I = moment of inertia of cylinder = 1/2mr² where r = radius of cylinder and v = translational velocity of cylinder. Its final potential energy U' = 0. Substituting these into the equation, we have
K' + U' = K + U
1/2Iω² + 1/2mv² + 0 = 0 + mgh
1/2(1/2mr²)ω² + 1/2mv² = mgh
1/4mr²ω² + 1/2mv² = mgh
1/4m(rω)² + 1/2mv² = mgh v = rω
1/4mv² + 1/2mv² = mgh
3/4mv² = mgh
v² = 4gh/3
v = √(4gh/3)
v = √(4 × 9.8 m/s² × 7.3 × 10⁻³ m/3)
v = (√286.16/3) m/s
v = (√95.39) m/s
v = 9.77 m/s
Help pleaseeeee!
If a force of 12 N is applied to a 4.0 kg object, what acceleration is produced?
Answer:
The answer is 3 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
From the question we have
[tex]a = \frac{12}{4} \\ [/tex]
We have the final answer as
3 m/s²Hope this helps you
